事象 $A,B,C$ が独立であることから、 \begin{align} P(A \cap B) &= P(A)P(B), \\ P(B \cap C) &= P(B)P(C), \\ P(C \cap A) &= P(C)P(A), \\ P(A \cap B \cap C) &= P(A)P(B)P(C) \end{align} が成り立つ。
(i) \begin{align} P(B) &= \frac{P(A \cap B)}{P(A)} \\ &= \frac{\frac{1}{5}}{\frac{1}{3}} \\ &= \frac{3}{5} \end{align}
(ii) \begin{align} P(A \cup C) &= P(A) + P(C) - P(A \cap C) \\ &= P(A) + P(C) - P(A)P(C) \\ &= P(A) + (1-P(A)) P(C) \\ \therefore \ \ P(C) &= \frac{P(A \cup C) - P(A)}{1 - P(A)} \\ &= \frac{\frac{3}{7} - \frac{1}{3}}{1 - \frac{1}{3}} \\ &= \frac{1}{7} \end{align}
(iii) \begin{align} P((A \cup B) \cap C) &= P((A \cap C) \cup (B \cap C)) \\ &= P(A \cap C) + P(B \cap C) - P(A \cap B \cap C) \\ &= P(A)P(C) + P(B)P(C) - P(A)P(B)P(C) \\ &= \frac{11}{105} \\ \therefore \ \ P(A \cup B \mid C) &= \frac{P((A \cup B) \cap C)}{P(C)} \\ &= \frac{\frac{11}{105}}{\frac{1}{7}} \\ &= \frac{11}{15} \end{align}